1
$\begingroup$

I was faced with the following question:

PART C – CHEMISTRY

Which one of the following properties is not shown by $\ce{NO}$?

  1. It combines with oxygen to form nitrogen dioxide

  2. Its bond order is $2.5$

  3. It is diamagnetic in the gaseous state

  4. It is a neutral oxide

I am confused about (3) & (4). I know $\ce{NO}$ reacts with water & oxygen to produce $\ce{HNO2}$, and then $\ce{NaOH}$ to produce $\ce{NaNO2, N2O, H2O}$. So, obviously it's acidic.

But there is one unpaired electron in the $\ce{N}$ atom. So, I see no reason for $\ce{NO}$ to be diamagnetic in gaseous state.

But, these questions do not have two correct answers. So, one of them has to be wrong. I am not sure which one.

$\endgroup$
1

1 Answer 1

4
$\begingroup$
  1. That's true.

  2. The bond order in $\ce{NO}$ is 2.5

  3. The oxidation state of $\ce N$ is +2, and this is a colourless, paramagnetic gas (i.e. it is a radical, or contains an unpaired electron). (source)

  4. Neutral oxides are those oxides which show neither basic nor acidic properties when they react with water. Examples include carbon monoxide $(\ce{CO})$ and nitrous oxide $(\ce{N2O})$ which are only slightly soluble in water, and nitric oxide $\mathbf{(NO)}$ which is appreciably soluble in cold water. (source)

So, option 3 is the correct answer as it is an inaccurate statement.

$\endgroup$
4
  • $\begingroup$ but NO reacts with NaOH. So, why isn't acidic? $\endgroup$ May 8, 2017 at 12:31
  • 1
    $\begingroup$ NO itself doesn't react with NaOH, so it is not acidic. It is only the downstream reaction products of NO with O2 that are acidic. $\endgroup$
    – hBy2Py
    May 8, 2017 at 13:15
  • $\begingroup$ Also, I take issue with that first citation. The oxidation of NO is second order in the NO concentration, so it is only a "quick" reaction when the concentration of NO is high. $\endgroup$
    – hBy2Py
    May 8, 2017 at 13:24
  • $\begingroup$ chemiday.com/en/reaction/3-1-0-8020 So is this wrong? $\endgroup$ May 8, 2017 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.